关于H¹(Ω)到Pₖ(Ω)的下界型投影类算子存在性的问询

Great question! The short answer is no—such an operator cannot exist. Let's break down why step by step, using key properties of functional analysis:

First, let's recap the definitions to make sure we're on the same page:

• $H^1(\Omega) = {v\in L^2(\Omega): \nabla v\in L^2(\Omega)2}$ is an infinite-dimensional Hilbert space (for any non-trivial domain $\Omega\subset\mathbb{R}^2$, it contains infinitely many linearly independent functions like $\sin(nx)\sin(ny)$ for $n\in\mathbb{N}$).
• $P_k(\Omega)$ is the set of polynomials on $\Omega$ with degree at most $k$, which is a finite-dimensional linear space (its dimension is $\binom{k+2}{2}$ for $\Omega\subset\mathbb{R}^2$).
• The norm in question is the standard $H^1$-norm: $|v|^2 = |v|_0^2 + |\nabla v|_0^2$, where $|\cdot|_0$ denotes the $L^2$-norm.

Why the operator can't exist #

Suppose for contradiction that there exists a linear operator $P:H^1(\Omega)\to P_k(\Omega)$ satisfying $|Pv| \geq C|v|$ for some constant $C>0$ independent of $v$. We can derive two contradictions from this assumption:

1. Injectivity vs. Dimension Mismatch

The inequality $|Pv| \geq C|v|$ immediately tells us $P$ is injective: if $Pv=0$, then $0 \geq C|v|$, so $|v|=0$, which means $v=0$ (since norms are positive-definite).

But an injective linear map from an infinite-dimensional space to a finite-dimensional space is impossible. Here's why:

• The image of $P$ is a subspace of $P_k(\Omega)$, which is finite-dimensional, so the image itself must be finite-dimensional.
• An injective linear map implies the domain ($H^1(\Omega)$) is isomorphic to its image. This would force $H^1(\Omega)$ to be finite-dimensional—but we know $H^1(\Omega)$ is infinite-dimensional. That's a direct contradiction.

2. Bounded Sequences in Finite vs. Infinite-Dimensional Spaces

We can also use properties of bounded sequences to reach a contradiction:

• Take an infinite orthonormal sequence ${e_n}$ in $H^1(\Omega)$ (such sequences exist because $H^1(\Omega)$ is an infinite-dimensional separable Hilbert space, assuming $\Omega$ is a nice domain like a bounded Lipschitz domain). For each $n$, $|e_n|=1$, so the inequality gives $|Pe_n| \geq C>0$.
• Since $P_k(\Omega)$ is finite-dimensional, every bounded sequence in it has a convergent subsequence (this is the Bolzano-Weierstrass theorem generalized to finite-dimensional normed spaces). Let ${Pe_{n_j}}$ be a convergent subsequence, with $Pe_{n_j} \to w \in P_k(\Omega)$.
• Orthonormal sequences in Hilbert spaces weakly converge to 0: for any $v\in H^1(\Omega)$, $\langle e_{n_j}, v\rangle \to 0$ as $j\to\infty$. Our operator $P$ is continuous (the inequality $|Pv - Pw| = |P(v-w)| \geq C|v-w|$ implies it's Lipschitz continuous), so weak convergence of $e_{n_j}\to 0$ implies $Pe_{n_j}\to P0=0$ weakly.
• But we also have $Pe_{n_j}\to w$ strongly, so $w$ must be 0. This contradicts $|Pe_{n_j}| \geq C>0$, since the norm of a convergent sequence converges to the norm of its limit (which would be 0).

Either way, we end up with a contradiction, so no such operator $P$ can exist.

内容的提问来源于stack exchange，提问作者yemino